If $(X_n)_{n\geq 1}$ is a submartingale and $S\leq T$ are bounded stopping times, then $\mathbb{E}(X_T)\geq\mathbb{E}(X_S)$.

I am quite new to conditional expectation, martingales etc. In our lecture the professor mentioned that the above can be proved by using the Doob Decomposition Theorem, however I don't quite see it. Could someone please explain this to me. Would be greatly appreciated, thank you in advance.

My attempt:

$$\begin{align*} \mathbb{E}(X_T)=\mathbb{E}(X_1+M_T+A_T) &=\mathbb{E}(X_1)+\mathbb{E}(M_T|\mathcal{F}_S)+\mathbb{E}(A_T)\\&=\mathbb{E}(X_1)+\mathbb{E}(\mathbb{E}(M_T|\mathcal{F}_S))+\mathbb{E}(A_T)\end{align*}$$ Then since $\mathbb{E}(M_T)=\mathbb{E}(M_0)$ for any martingale $M_n$ and any bounded stopping time T, we have $\mathbb{E}(\mathbb{E}(M_T|\mathcal{F}_S))=\mathbb{E}(M_0)$, so $$\begin{align*}&=\mathbb{E}(X_1)+\mathbb{E}(M_0)+\mathbb{E}(A_T)\end{align*}$$ Now since $M_S$ is also a martingale and S is also a bounded stopping time, we again have $\mathbb{E}(M_0)=\mathbb{E}(M_S)$, $$\begin{align*}&=\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_T)\end{align*}$$ Finally, since according to Doob's Decomposition theorem $(A_n)_{n\geq 0}$ is an increasing predictable process, $$\begin{align*}\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_T)&\geq\mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_S)\\ &=\mathbb{E}(X_S)\end{align*}$$

  • 1
    $\begingroup$ Do you know that $\mathbb{E}(M_T) = \mathbb{E}(M_0)$ for any martingale $(M_n)_n$ and any bounded stopping time $T$? $\endgroup$
    – saz
    Oct 22, 2017 at 17:43
  • $\begingroup$ Ah yes! Thank you for the tip. Above I have shown my attempt to prove it using this. Does it make sense what I have done? $\endgroup$
    – math189925
    Oct 23, 2017 at 21:38
  • $\begingroup$ You don't have to condiition on $\mathcal{F}_S$. Just note that $\mathbb{E}(M_T) = \mathbb{E}(M_0) = \mathbb{E}(M_S)$ and so $$\mathbb{E}(X_1+M_T+A_T) = \mathbb{E}(X_1)+\mathbb{E}(M_T) + \mathbb{E}(A_T) = \mathbb{E}(X_1)+\mathbb{E}(M_S)+\mathbb{E}(A_T).$$ $\endgroup$
    – saz
    Oct 24, 2017 at 5:24
  • $\begingroup$ Ok i see. Thank you for your help :) $\endgroup$
    – math189925
    Oct 24, 2017 at 10:35
  • $\begingroup$ You are welcome. $\endgroup$
    – saz
    Oct 24, 2017 at 10:59

1 Answer 1


Suppose $S\leq T \leq t$. By Doob Decomposition, there exists $(M_n)_n$ a martingale and $(A_n)_n$ increasing and predictable such that $\forall n, X_n=M_n+A_n$. Then $$\begin{aligned}X_S &= M_S + A_S = E(M_t|\mathcal F_S) + A_S \quad \text{by lemma 1 } \\ &= E(M_t + A_S|\mathcal F_S)\\ &\leq E(M_t + A_T|\mathcal F_S) = E( E(M_t + A_T |\mathcal F_T)|\mathcal F_S) \quad \text{since } F_S \subset F_T\\ &= E(M_T + A_T | \mathcal F_S) = E(X_T|\mathcal F_S) \end{aligned}$$

Taking expectations on both sides, $E(X_S)\leq E(X_T)$.

Lemma 1: Let $(X_n)_n$ be a martingale and $\tau$ a bounded stopping time. Let $t\in \mathbb N$ such that $\tau \leq t$. Then $X_\tau = E(X_t|\mathcal F_\tau)$.

Proof: It suffices to prove $$\forall A\in \mathcal F_\tau, E(\mathbb 1_A X_t) = E(\mathbb 1_A X_\tau)$$ Recall (or prove) that $X_\tau$ is $\mathcal F_\tau$ measurable. $$\begin{aligned}E(\mathbb 1_A X_\tau) &= E(\mathbb 1_A \sum_{k=0}^t X_k \mathbb 1_{\tau = k}) = \sum_{k=0}^t E(\mathbb 1_{A\cap (\tau = k)} E(X_t|\mathcal F_k)) \\ &= \sum_{k=0}^t E( E(\mathbb 1_{A\cap (\tau = k)}X_t|\mathcal F_k)) \\ &= \sum_{k=0}^t E(\mathbb 1_{A\cap (\tau = k)}X_t) = E(\mathbb 1_A X_t) \end{aligned}$$


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